Convex cocompactness in pseudo-Riemannian hyperbolic spaces

Jeffrey Danciger; François Guéritaud; Fanny Kassel

doi:10.1007/s10711-017-0294-1

Geometriae Dedicata

February 2018, Volume 192, Issue 1, pp 87–126 | Cite as

Convex cocompactness in pseudo-Riemannian hyperbolic spaces

Authors
Authors and affiliations

Jeffrey Danciger
François Guéritaud
Fanny KasselEmail author

Original Paper

First Online: 04 December 2017

Received: 31 January 2017

Accepted: 13 October 2017

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Abstract

Anosov representations of word hyperbolic groups into higher-rank semisimple Lie groups are representations with finite kernel and discrete image that have strong analogies with convex cocompact representations into rank-one Lie groups. However, the most naive analogy fails: generically, Anosov representations do not act properly and cocompactly on a convex set in the associated Riemannian symmetric space. We study representations into projective indefinite orthogonal groups $\mathrm {PO}(p,q)$ by considering their action on the associated pseudo-Riemannian hyperbolic space $\mathbb {H}^{p,q-1}$ in place of the Riemannian symmetric space. Following work of Barbot and Mérigot in anti-de Sitter geometry, we find an intimate connection between Anosov representations and a natural notion of convex cocompactness in this setting.

Keywords

Convex cocompact groups Pseudo-Riemannian hyperbolic spaces Real projective geometry Anosov representations Right-angled Coxeter groups

Mathematics Subject Classification

20F55 22E40 52A20 53C50 57S30

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Notes

Acknowledgements

We are grateful to Yves Benoist and Anna Wienhard for motivating comments and questions, and for their encouragement. We also thank Vivien Ripoll for interesting discussions on limit sets of Coxeter groups, the referee for useful suggestions, and Jean-Philippe Burelle, Virginie Charette and Son Lam Ho for pointing out a subtlety in Sect. 4. The main results, examples, and ideas of proofs in this paper were presented by the third-named author in June 2016 at the conference Geometries, Surfaces and Representations of Fundamental Groups in honor of Bill Goldman; we would like to thank the organizers for a very interesting and enjoyable conference. Finally, we thank Bill Goldman for being a constant source of inspiration and encouragement to us and many others in the field.

Appendix A: Connectedness in the space of unordered tuples

The following general statement, on which Proposition 1.10 relies, is probably well known. We provide a proof for the reader’s convenience.

Fact A.1

Let $\Lambda $ be a connected Hausdorff topological space. For $k\ge 1$, the space $\Lambda ^{(k)}$ of unordered k-tuples of pairwise distinct points of $\Lambda $ is also connected (for the restriction of the product topology).

Given a finite subset X of $\Lambda $ and a point $x\in \Lambda \smallsetminus X$, we denote by $\Lambda _x^X$ the connected component of $\Lambda \smallsetminus X$ containing x. Since $\Lambda $ is Hausdorff, its finite subsets are closed, and so $\Lambda _x^X$ is an open subset of $\Lambda $ and its closure $\overline{\Lambda _x^X}$ is contained in $\Lambda _x^X\cup X$.

Lemma A.2

Let $\Lambda $ be a connected topological space with closed singletons. For any $k\ge 1$ and $\{x_0, \dots , x_k\}\in \Lambda ^{(k+1)}$, there exists $0\le i_0 < k$ such that the unordered k-tuples $\{x_0, \dots , x_k\} \smallsetminus \{x_{i_0}\} $ and $\{x_0, \dots , x_{k-1}\}$ belong to the same connected component of $\Lambda ^{(k)}$.

Proof

For $0\le i<k$, let $X_i:=\{x_0, \dots , x_{k-1}\}\smallsetminus \{x_i\}$. It is sufficient to prove the existence of $0\le i_0 < k$ such that $x_{i_0}$ and $x_k$ belong to the same connected component of $\Lambda \smallsetminus X_{i_0}$, i.e. $x_{i_0}\in \Lambda _{x_k}^{X_{i_0}}$. We have

$$\begin{aligned} \overline{\bigcap _{0\le i<k} \Lambda _{x_k}^{X_i}} \subset \bigcap _{0\le i<k} \overline{\Lambda _{x_k}^{X_i}} \subset \bigcap _{0\le i <k} \left( \Lambda _{x_k}^{X_i} \cup X_i\right) . \end{aligned}$$

(A.1)

Suppose by contradiction that $x_i\notin \Lambda _{x_k}^{X_i}$ for all $0\le i<k$: then $x_i\notin \Lambda _{x_k}^{X_i} \cup X_i$, so the right-hand intersection in (A.1) is disjoint from all $X_i$ and can be rewritten $\bigcap _{0\le i <k} \Lambda _{x_k}^{X_i}$. This set is therefore open and [by (A.1)] closed, and contains $x_k$ but no other $x_i$, contradicting the fact that $\Lambda $ is connected. $\square $

Lemma A.3

Let $\Lambda $ be a connected topological space with closed singletons. For any $k\ge 1$ and $\{x_0, \dots , x_k\}\in \Lambda ^{(k+1)}$, there exists $1\le j_0 \le k$ such that $x_i \in \Lambda _{x_0}^{\{x_{j_0}\}}$ for all $i\in \{1,\dots ,k\}\smallsetminus \{ j_0\}$.

We call this property $\mathrm {H}_k$, or $\mathrm {H}_k(x_0, \{x_1,\dots , x_k\})$ to be specific.

Proof

We argue by induction. Property $\mathrm {H}_1$ is vacuously true. Assuming $\mathrm {H}_{k-1}$ where $k\ge 2$, let us prove $\mathrm {H}_k$ by contradiction. We have

$$\begin{aligned} \overline{\bigcap _{1\le j \le k} \Lambda _{x_0}^{\{x_j\}}} \subset \bigcap _{1\le j \le k} \overline{\Lambda _{x_0}^{\{x_j\}}} \subset \bigcap _{1\le j \le k} \left( \Lambda _{x_0}^{\{x_j\}} \cup \{x_j\}\right) . \end{aligned}$$

(A.2)

Suppose $\mathrm {H}_k(x_0, \{x_1,\dots , x_k\})$ fails: that is, for all $1\le j \le k$,

$$\begin{aligned} \{x_1,\dots ,x_k\}\smallsetminus \{ x_j\} \not \subset \Lambda _{x_0}^{\{x_{j}\}}. \end{aligned}$$

We claim that the right member of (A.2) then cannot contain any $x_i$ for $1\le i \le k$: indeed that would imply $x_i\in \Lambda _{x_0}^{\{x_j\}}$ for all $j\in \{1,\dots ,k \} \smallsetminus \{i\}$, hence the above relationship would yield

$$\begin{aligned} \{x_1,\dots ,x_k\}\smallsetminus \{x_i, x_j\} \not \subset \Lambda _{x_0}^{\{x_{j}\}} \end{aligned}$$

for all $j\in \{1,\dots , k\}\smallsetminus \{i\}$, contradicting $\mathrm {H}_{k-1}(x_0, \{x_1,\dots , x_k\}\smallsetminus \{x_i\})$.

Therefore the right-hand side of (A.2) can be written $\bigcap _{1\le j \le k} \Lambda _{x_0}^{\{x_j\}}$, which by (A.2) turns out to be closed. It is also open, and contains $x_0$ but no other $x_i$: this contradicts connectedness of $\Lambda $. Therefore $\mathrm {H}_k$ holds. $\square $

Proof of Fact A.1

We argue by induction on k to prove that $\Lambda ^{(k)}$ is connected for any connected topological space $\Lambda $ with closed singletons. The case $k=1$ is obviously true. For $k\ge 2$, suppose that $(\Lambda ')^{(k-1)}$ is connected for any connected $\Lambda '$ with closed singletons, and let us prove that $\Lambda ^{(k)}$ is connected for any connected $\Lambda $ with closed singletons.

Consider $\{x_0, \dots , x_k\}\in \Lambda ^{(k+1)}$. By Lemma A.3, up to exchanging the labels $j_0$ and k, we have $x_i \in \Lambda _{x_0}^{\{x_k\}}$ for all $1\le i\le k-1$, i.e. all points $x_0, \dots , x_{k-1}$ belong to the same connected component $\Lambda '$ of $\Lambda \smallsetminus \{x_k\}$. Since $\Lambda '^{(k-1)}$ is connected, all $(k-1)$-tuples $\{x_0, \dots , x_{k-1}\}\smallsetminus \{x_i\}$ for $0\le i < k$ belong to the same connected component of $(\Lambda \smallsetminus \{x_k\})^{(k-1)}$, and so all k-tuples $\{x_0, \dots , x_k\}\smallsetminus \{x_i\}$ for $0\le i < k$ belong to the same component of $\Lambda ^{(k)}$. But by Lemma A.2 one of these k-tuples (for $i=i_0$) belongs to the same component as $\{x_0, \dots , x_{k-1}\}$. Therefore all k-tuples contained in $\{x_0, \dots , x_k\}$ belong to the same component of $\Lambda ^{(k)}$. This is true for all $\{x_0, \dots , x_k\}\in \Lambda ^{(k+1)}$, hence $\Lambda ^{(k)}$ is connected. $\square $

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Authors and Affiliations

Jeffrey Danciger
- 1
François Guéritaud
- 2
Fanny Kassel
- 3
Email author

1.Department of MathematicsThe University of Texas at AustinAustinUSA
2.Laboratoire Paul PainlevéCNRS and Université Lille 1Villeneuve d’Ascq CedexFrance
3.Laboratoire Alexander GrothendieckCNRS and Institut des Hautes Études ScientifiquesBures-sur-YvetteFrance

Convex cocompactness in pseudo-Riemannian hyperbolic spaces

Abstract

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